If #(7x^2)/(2x-3)#, what are the points of inflection, concavity and critical points?

1 Answer
Apr 26, 2017

There is a local maximum at #(0,0)# and a local minimum at #(3,21)#
No inflection point.
The function is concave in the interval #(-oo,3/2)# and convex in the interval #(3/2,+oo)#

Explanation:

We need

#(u/v)'=(u'v-uv')/(v^2)#

We calculate the first and second derivatives

Let #f(x)=(7x^2)/(2x-3)#

#u=7x^2#, #=>#, #u'=14x#

#v=2x-3#, #=>#, #v'=2#

Therefore,

#f'(x)=(14x(2x-3)-7x^2(2))/(2x-3)^2#

#=(28x^2-42x-14x^2)/(2x-3)^2#

#=(14x^2-42x)/(2x-3)^2#

#=(14x(x-3))/(2x-3)^2#

The critical points are when #x=0# and #x=3#

We can build a chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##0##color(white)(aaaaaaaa)##3/2##color(white)(aaaaaaa)##3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x##color(white)(aaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-3##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f'(x)##color(white)(aaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##↗##color(white)(aaaa)##↘##color(white)(aaaa)##||##color(white)(aaaa)##↘##color(white)(aaaa)##↗#

There is a local maximum at #(0,0)# and a local minimum at #(3,21)#

Now, we determine the second derivative

#u=(14x^2-42x)#, #=>#, #u'=28x-42#

#v=(2x-3)^2#, #=>#, #v'=4(2x-3)#

#f''(x)=((28x-42)(2x-3)^2-4(14x^2-42x)(2x-3))/(2x-3)^4#

#=((2x-3)(56x^2-84x-84x+126-56x^2+168x))/(2x-3)^4#

#=126/(2x-3)^3#

There is no inflexion point.

We can build a chart

#color(white)(aaaa)##Interval##color(white)(aaaa)##(-oo,3/2)##color(white)(aaaa)##(3/2,+oo)#

#color(white)(aaaa)##f''(x)##color(white)(aaaaaaaaaa)##-##color(white)(aaaaaaaaaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaaaaaa)##nn##color(white)(aaaaaaaaaaa)##uu#

The function is concave in the interval #(-oo,3/2)# and convex in the interval #(3/2,+oo)#

graph{7x^2/(2x-3) [-66.14, 65.5, -22.55, 43.3]}